九度OJ1002

题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：15412

解决：3984

题目描述：

    Grading hundreds of thousands of Graduate Entrance Exams is a hard work. It is even harder to design a process to make the results as fair as possible. One way is to assign each exam problem to 3 independent experts. If they do not agree to each other, a judge is invited to make the final decision. Now you are asked to write a program to help this process.
    For each problem, there is a full-mark P and a tolerance T(<P) given. The grading rules are:
    • A problem will first be assigned to 2 experts, to obtain G1 and G2. If the difference is within the tolerance, that is, if |G1 - G2| ≤ T, this problem's grade will be the average of G1 and G2.
    • If the difference exceeds T, the 3rd expert will give G3.
    • If G3 is within the tolerance with either G1 or G2, but NOT both, then this problem's grade will be the average of G3 and the closest grade.
    • If G3 is within the tolerance with both G1 and G2, then this problem's grade will be the maximum of the three grades.
    • If G3 is within the tolerance with neither G1 nor G2, a judge will give the final grade GJ.

输入：

    Each input file may contain more than one test case.
    Each case occupies a line containing six positive integers: P, T, G1, G2, G3, and GJ, as described in the problem. It is guaranteed that all the grades are valid, that is, in the interval [0, P].

输出：

    For each test case you should output the final grade of the problem in a line. The answer must be accurate to 1 decimal place.

样例输入：

20 2 15 13 10 18

样例输出：

14.0

来源：
2011年浙江大学计算机及软件工程研究生机试真题
答疑：
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CODE:

#include <iostream>#include <iomanip>#include <cmath>using namespace std;int main(){    double P,T,G1,G2,G3,GJ;    double sum;    while(cin >> P >> T >> G1 >> G2 >> G3 >> GJ)    {        if(fabs(G1-G2)<=T)            sum=(G1+G2)/2;            else if(fabs(G3-G1)<=T&&fabs(G3-G2)<=T)                sum=(G1+G2+G3)/3;        else if(fabs(G3-G1)<=T)            sum=(G1+G3)/2;        else if(fabs(G3-G2)<=T)            sum=(G2+G3)/2;        else            sum=GJ;        cout << setiosflags(ios::fixed) << setprecision(1) << sum << endl;    }    return 0;}